TITLE: Probability a die will come up 6 at least twice in twelve rolls
QUESTION [6 upvotes]: What is the probability of rolling at least two 6's on twelve rolls of a fair 6-sided die.
I am using the complement to solve the question
$$1-\frac {5^{12}+(_{12}C_ 1)5^{11}}{6^{12}}$$
$5^{12}$ is the probability of not rolling a six
$6^{12}$ is the sample space
My question is why does the binomial coefficient have to be multiplied $5^{11}$ where the binomial coefficient is a "6" on the die.
This method I find difficult but intriguing.

REPLY [5 votes]: $\left(\frac56\right)^{12}$ is the probability that no sixes are rolled in 12 rolls.
$\frac16\left(\frac56\right)^{11}$ is the probability that a six is rolled on the first roll and then none are rolled in the remaining $11$ rolls. Now we have to count the possibilities that each of the $12$ dice could be six and the rest not. Therefore, we multiply by $\binom{12}{1}=12$.
Thus, the probability of rolling at least $2$ sixes would be the complement of the sum of these probabilities:
$$
1-\left(\frac56\right)^{12}-\binom{12}{1}\frac16\left(\frac56\right)^{11}
$$

If we want to count the number of ways to roll exactly $k$ sixes, this would be $\binom{12}{k}\left(\frac16\right)^k\left(\frac56\right)^{11-k}$ since $\binom{12}{k}$ is the number of ways to choose the dice on which the sixes are rolled.